Leetcode 1344: Angle Between Hands of a Clock
Given an hour and minute value, return the smaller angle (in degrees) formed between the hour and minute hands of a clock.
Problem
Approach
Steps
Complexity
Input: The input consists of two integers representing the hour (1 to 12) and the minute (0 to 59).
Example: hour = 6, minutes = 45
Constraints:
• 1 <= hour <= 12
• 0 <= minutes <= 59
Output: Return the smaller angle in degrees between the hour and minute hands of a clock.
Example: For hour = 6, minutes = 45, the output will be 22.5.
Constraints:
• The answer should be within 10^-5 of the actual value.
Goal: Calculate the angle between the hour and minute hands of a clock and return the smaller angle.
Steps:
• 1. Calculate the angle of the minute hand based on the number of minutes.
• 2. Calculate the angle of the hour hand, adjusting for the current minute.
• 3. Find the absolute difference between the two angles.
• 4. Return the smaller angle (either the absolute difference or 360 - difference).
Goal: The solution should handle all valid clock inputs efficiently.
Steps:
• The hour value is between 1 and 12.
• The minute value is between 0 and 59.
Assumptions:
• The given time values are always valid.
• The hour value is always between 1 and 12, and the minute value is between 0 and 59.
• Input: Example 1: hour = 6, minutes = 45
• Explanation: At 6:45, the minute hand is at the 9th hour mark, and the hour hand is at 6:45, creating an angle of 22.5 degrees.
• Input: Example 2: hour = 9, minutes = 15
• Explanation: At 9:15, the minute hand is at the 3rd hour mark, and the hour hand is at 9:15, forming an angle of 52.5 degrees.
• Input: Example 3: hour = 10, minutes = 0
• Explanation: At 10:00, the angle between the two hands is exactly 60 degrees, as the minute hand is at 12 and the hour hand is at 10.
Approach: To find the smaller angle between the hour and minute hands, we calculate the positions of both hands and then compute the absolute difference between them.
Observations:
• The minute hand moves 6 degrees per minute, and the hour hand moves 30 degrees per hour, so we can use these values to calculate the angles.
• The key challenge is to adjust for the fact that the hour hand also moves as the minutes progress. This needs to be factored into the calculation.
Steps:
• 1. Compute the angle of the minute hand using the formula: 6 * minutes.
• 2. Compute the angle of the hour hand using the formula: 30 * (hour % 12) + (minutes / 60) * 30.
• 3. Calculate the absolute difference between the two angles.
• 4. Return the smaller of the absolute difference and 360 - the difference.
Empty Inputs:
• There are no empty inputs since hour and minute are always valid values.
Large Inputs:
• There are no large inputs as the input values are bounded by 12 for the hour and 59 for the minutes.
Special Values:
• When minutes = 0, the angle between the hands will be a multiple of 30 degrees.
Constraints:
• Ensure that all calculations are performed correctly for boundary values like hour = 12 and minutes = 0.
double angleClock(int h, int minutes) {
double mn = 360 *(double)minutes/60;
double hrs = 360 * ((double)(h == 12? 0: h) / 12) + 30* ((double)minutes/60);
cout << mn << " " << hrs;
return min(abs(mn - hrs), 360 - (abs(mn - hrs)));
}
1 : Function Declaration
double angleClock(int h, int minutes) {
Declares a function that calculates the angle between the hour and minute hands of a clock, given the hour and minute inputs.
2 : Minute Calculation
double mn = 360 *(double)minutes/60;
Calculates the angle of the minute hand relative to the 12 o'clock position. Each minute represents 6 degrees.
3 : Hour Calculation
double hrs = 360 * ((double)(h == 12? 0: h) / 12) + 30* ((double)minutes/60);
Calculates the angle of the hour hand, considering both the hour and fractional contribution of the minutes. Accounts for `h == 12` by resetting to 0.
4 : Angle Calculation
return min(abs(mn - hrs), 360 - (abs(mn - hrs)));
Returns the smaller angle between the hour and minute hands by calculating both the direct and complementary angles.
Best Case: O(1)
Average Case: O(1)
Worst Case: O(1)
Description: The time complexity is constant since we only perform a few arithmetic operations.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant as we only need a few variables to store intermediate results.
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